dc step up circuit

Question

I'm looking to power my old spy gear trackr with rechargeable batteries,but I need to boost a 3.6v 1200ma battery to 9v @1200ma

Its ni-mh and a ni-cd 2.4v @ 800ma to 6v @ 800ma I know that step-up circuits reduce the amount of current it outputs so how many Ma should the battery be to compensate for the reduction and could someone also help me understand boost circuits so I can create my own in the future

-thanks Gabriel wells

jw0752 · Accepted Answer

Hi Gabe,

You can do a rough calculation of the maximum you can expect on a step up by multiplying the amps times the volts of the input and it will be equal to the amps times the volts of the output. For example if you have 3.6v and 1200mA ( which is 1.2A) then we get 3.6V X 1.2A = 4.32VA If we step up to 9volts we just divide 4.32VA by the 9 volts and get .48A or in other words 480mA. In real life there is always some loss in the change over so these numbers are going to be too high. Perhaps we might realistically expect to get 9 volts 400mA.

Here is a link to a circuit that can perhaps make the boost that you are looking for. If not there are many different ones listed too.

http://www.ebay.com/itm/DC-DC-Adjustable-Step-up-boost-Power-Converter-Module-XL6009-Replace-LM2577-/181608388242?pt=LH_DefaultDomain_0&hash=item2a48b41692

I would check the internet for information on switching and boost circuits if you want more information on how they are designed as it isn't simple.

John

dc step up circuit

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